The Runtime Complexity (full) of the given CpxTRS could be proven to be BOUNDS(n^1, INF).

(0) Obligation:

Runtime Complexity TRS:
The TRS R consists of the following rules:

intlist(nil) → nil
int(s(x), 0) → nil
int(x, x) → cons(x, nil)
intlist(cons(x, y)) → cons(s(x), intlist(y))
int(s(x), s(y)) → intlist(int(x, y))
int(0, s(y)) → cons(0, int(s(0), s(y)))
intlist(cons(x, nil)) → cons(s(x), nil)

Rewrite Strategy: FULL

(1) CpxTrsToCpxRelTrsProof (BOTH BOUNDS(ID, ID) transformation)

Transformed TRS to relative TRS where S is empty.

(2) Obligation:

Runtime Complexity Relative TRS:
The TRS R consists of the following rules:

intlist(nil) → nil
int(s(x), 0) → nil
int(x, x) → cons(x, nil)
intlist(cons(x, y)) → cons(s(x), intlist(y))
int(s(x), s(y)) → intlist(int(x, y))
int(0, s(y)) → cons(0, int(s(0), s(y)))
intlist(cons(x, nil)) → cons(s(x), nil)

S is empty.
Rewrite Strategy: FULL

(3) DecreasingLoopProof (EQUIVALENT transformation)

The following loop(s) give(s) rise to the lower bound Ω(n¹):
The rewrite sequence
intlist(cons(x, y)) →⁺ cons(s(x), intlist(y))
gives rise to a decreasing loop by considering the right hand sides subterm at position [1].
The pumping substitution is [y / cons(x, y)].
The result substitution is [ ].

(4) BOUNDS(n^1, INF)